/*
 * File: one.c
 * Author: Antonio Pessoa
 * Date: 17/05/2012
 * Description: If we list all the natural numbers below 10 that are multiples
 *              of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is
 *              23. Find the sum of all the multiples of 3 or 5 below 1000.
 */

#include <stdio.h>

#define LIMIT 1000

void one() {
    int sum = 0;
    
    for (int i = 1; i < LIMIT; i++) {
        if ((i % 3) == 0 || (i % 5) == 0)
            sum += i;
    }
    
    printf("\nTotal: %d\n\n", sum);
}
